It is a classical result that if $0^{\sharp}$ exists, then there is a model of $ZFC$ in which there is a $\Delta^1_3$ well ordering of reals but no $\Delta^1_2$-well ordering.

My question is: Is $0^{\sharp}$ necessary to prove this? Or does the statement $\mathbf{CON}(ZFC+\mbox{there is a }\Delta^1_3\mbox{ well ordering of reals but no }\Delta^1_2 )$ implies the consistency of the existence of $0^{\sharp}$?

it is shown that it is equiconsistent with $\mathsf{ZFC}$ to have a boldface $\Delta^1_3$ well-ordering and Martin's axiom. But the existence of boldface $\Delta^1_2$ well-orderings implies that the reals are the reals of $L[r]$ for some real $r$ (this is a result of Mansfield), and this fails under Martin's axiom.

(The boldface in Leo's result can be made lightface at the cost of more complicated coding techniques. This was shown by Sy Friedman, see his book on "Class forcing". If we do not care about also having $\mathsf{MA}$, Leo's paper also shows how to obtain from $\mathsf{ZFC}$ models with lightface $\Delta^1_3$ well-orderings.)

Going beyond $\mathsf{MA}$, recently, Sy and I proved that if $\mathsf{BPFA}$ holds and $\omega_1=\omega_1^L$, then there is a lightface $\Delta^1_3$ well-ordering of $\mathbb R$. Curiously, it is still open whether starting with $L$ and using the standard forcing for $\mathsf{MA}$ ($+2^\omega=\omega_2$) results in a model with a definable well-ordering of $\mathbb R$.

On the other hand, there is a connection with sharps: Assume that there are no inner models with a strong cardinal (or there is a measurable, but there are no inner models with a Woodin cardinal). If all reals have sharps, and there is a $\Delta^1_3(r)$ well-ordering (for some real $r$), then the reals are the reals of the core model $K_r$. This was first noted by Welch.